I don't get why for $u$ substitution they sub the upper and lower bounds
into $u$ from the original function to find the new upper and lower
bounds with the function $u$.
[...]
My question is what or how does plugging your old lower and upper
bound values into $u$ give you the new values of your new function that's
expressed as $u$...
The blunt answer is "That's what the change of variables theorem says."
But here's a more conceptual and notational explanation. The notation $\int_a^b f(x)\, dx$ connotes "integrating from $x = a$ to $x = b$". To emphasize this, let's write
$$
\int_{x=a}^{x=b} f(x)\, dx.
$$
Now suppose you want to evaluate
$$
\int_{x=a}^{x=b} f\bigl(g(x)\bigr) g'(x)\, dx.
$$
If you make the substitution $u = g(x)$, then $du = g'(x)\, dx$ ("by the chain rule"). If $F$ denotes an antiderivative of $f$, the preceding becomes
$$
\int_{x=a}^{x=b} f(u)\, du = F(u) \Big|_{x=a}^{x=b}.
$$
Now, it should be notationally clear that setting $u = a$ and $u = b$ does not (in general) "give the right answer": Those are not the limits specified in the original integral.
To proceed, you have two choices:
Undo the original substitution by setting $u = g(x)$, and then plug in $x = b$ and $x = a$.
Find the "new limits of integration", $u = g(a)$ and $u = g(b)$, by plugging the "old" limits $x = a$ and $x = b$ into the substitution $u = g(x)$.
The second makes notational sense because "when $x = a$, we have $u = g(a)$" and "when $x = b$, we have $u = g(b)$". It should be procedurally clear the two methods are mathematically equivalent. Computationally, the second is usually less work (as both prior respondents note); the first amounts to writing something down, then erasing it.
In symbols, either approach gives
\begin{align*}
\int_{x=a}^{x=b} f\bigl(g(x)\bigr) g'(x)\, dx
&= \int_{x=a}^{x=b} f(u)\, du && \text{Substitute $u = g(x)$;} \\
&= F(u) \Big|_{x=a}^{x=b} && \text{Antidifferentiate;} \\
&= F(u) \Big|_{u=g(a)}^{u=g(b)} && \text{Change limits;} \\
&= F\bigl(g(b)\bigr) - F\bigl(g(a)\bigr) && \text{Plug in;} \\
&= \int_{u=g(a)}^{u=g(b)} f(u)\, du. && \text{Reinterpret the fundamental theorem.}
\end{align*}
So much for the explanation; what about Real Life? The notation
$$
\int_{x=a}^{x=b} f(x)\, dx
$$
is redundant, and in practice, out of laziness^H^H^H elegance, we fall back on $\int_a^b f(x)\, dx$ (sometimes to the confusion of calculus students). But when you're learning substitution the first time, it may help to write in the variable corresponding to the numerical limits, as in:
\begin{align*}
\int_{x=0}^{x=\pi/4} -2\cos^{2}(2x) \sin(2x)\, dx
&= \int_{x=0}^{x=\pi/4} u^{2}\, du && u = \cos(2x),\quad du = -2\sin(2x)\, dx; \\
&= \int_{u=1}^{u=0} u^{2}\, du && \text{When $x = 0$, $u = 1$; when $x = \pi/4$, $u = 0$.}
\end{align*}
Solve for $x$:
$$t^2 = \frac{x}{x+3} = 1 - \frac{3}{x+3} \implies x = \frac{3}{1-t^2} - 3$$
then we have
$$dx = \frac{6t}{(1-t^2)^2}dt$$
and plugging in to the integral gets us
$$ \int_0^{\frac{1}{2}} \left(\frac{1-t^2}{3}\right)\cdot (t) \cdot \left(\frac{6t}{(1-t^2)^2}\right)dt = 2\int_0^{\frac{1}{2}} \frac{t^2}{1-t^2}dt = \int_0^{\frac{1}{2}} \frac{2}{1-t^2}-2dt$$
$$ = \int_0^{\frac{1}{2}} \frac{1}{1+t}+\frac{1}{1-t}-2dt = \log\left(\frac{1+t}{1-t}\right)-2t\Biggr|_0^{\frac{1}{2}} = \log(3)-1$$
Best Answer
Your question is incomplete without showing how you might even "compute the rest of the new integral in terms of $u$". I tried to point this out in the comments, but you haven't done so, so I will have to guess what you're thinking.
It's possible you mean something like:
$$u = x^2 \implies x = \begin{cases}\sqrt{u}, & \text{ if } x \geq 0 \\ -\sqrt{u}, & \text{ otherwise}\end{cases} \\ dx = \begin{cases}\frac{1}{2\sqrt{u}}\,du, & \text{ if } x \geq 0\\ -\frac{1}{2\sqrt{u}}\,du, & \text{ otherwise}\end{cases}$$
In other words, this substitution gives you
$$\int_{-1}^1 x^2 \,dx = \int_{-1}^0 x^2\,dx + \int_0^1 x^2\,dx = \int_{1}^0 u\left(-\frac{du}{2\sqrt{u}}\right) + \int_0^1 u\left(\frac{du}{2\sqrt{u}}\right)$$ which if we want, we can recombine to $$\int_0^1\sqrt{u}\,du$$ but we still don't get $0$.